Question

5. Which one of the following did not prevail during the period of Chandragupta-1|?
(a) Various branches of science were well developed.
(b) Education was available to people of all the classes in the society.
(c) Agriculture was the main occupation of the people.
(d) Most of the people including the king followed Shaivism and Vaishnavism.​

Answer 1

$\large{\underline{\underline{\mathfrak{\green{\bf{\:QUESTION:-}}}}}}$

1) when two resistors of resistance

R1, and R2 are connected in ll the

het resistance is 3 ohm.

when connected in series, its value is 16,

calculate the values of R, and R2.

$\large{\underline{\underline{\mathfrak{\red{\bf{\:ANSWER:-}}}}}}$

$\large{\underline{\:GIVEN\:HERE:-}}$

when two resistors of resistance

R1, and R2 are connected in ll the

het resistance is 3 ohm.

when connected in series, its value is 16 .

$\bold{\underline{\:FIND\:HERE:-}}$

Calculate the Value of resistance R1 and R2 .

$\large{\underline{\underline{\mathfrak{\red{\bf{\:EXPLANATION:-}}}}}}$

we know,

When two resistance R1 and R2 are connected in parallel

So, Total resistance will be

$\bold{\boxed{\boxed{\:\frac{1}{R}\:=\:\frac{1}{R1}\:+\frac{1}{R2}}}}$

when two resistance R1 and R2 are connected in series .

So, Total resistance will be

$\bold{\boxed{\boxed{\:R\:=\:R1\:+\:R2}}}$

_______________________

A/C to question ,

$\frac{1}{R1}\:+\frac{1}{R2}\:=\frac{1}{3}$

$\leadsto\:(\frac{R1+R2)}{R1R2})\:=\frac{1}{3}$

$\leadsto\:R1R2\:=\:3\times\:(R1+R2)$.....(1)

And,

$\:R1\:+\:R2\:=\:16$.....(1)

Keep value by (2) , in equation (2)

$\leadsto\:R1R2\:=\:3\times\:16$

$\bold{\boxed{\boxed{\:R1R2\:=\:48}}}$....(3)

But, we know,

$\:(a-b)\:=\sqrt{(a+b)^2-4ab}$

So,

$\bold{\boxed{\boxed{(R1-R2)\:=\sqrt{(R1+R2)^2-4R1R2}}}}$

keep value by (1) and (3)

$\leadsto\:(R1-R2)\:=\sqrt{(16)^2-4\times\:48}$

$\leadsto\:(R1-R2)\:=\sqrt{256-\:192}$

$\leadsto\:(R1-R2)\:=\sqrt{64}$

$\bold{\boxed{\boxed{\:(R1-R2)\:=\:8}}}$....(4)

Now, we have

R1 + R2 = 16

R1 - R2 = 8

__________________

Add this, we find

$\leadsto\:2R1\:=\:24$

$\leadsto\:R1\:=\cancel{\frac{24}{2}}$

$\bold{\boxed{\boxed{\:R1\:=\:12}}}$

keep value of R1 in (1),

$\leadsto\:(12+R2)\:=\:16$

$\leadsto\:R2\:=\:16-12$

$\bold{\boxed{\boxed{R2\:=\:4}}}$

$\large{\underline{\underline{\mathfrak{\bf{\:ANSWER:-}}}}}$

Resistance of R1 = 12 ohm

Resistance of R2 = 4 ohm

___________________

$\large{\underline{\underline{\mathfrak{\bf{\:ANSWER\:VERIFICATION:-}}}}}$

Case(1):- when R1 and R2 in | | .

Total Resistance will be

$\implies\frac{R1R2}{(R1+R2)}\:=\:16$

keep value of R1 and R2,

$\implies\frac{12×4}{(12+4)}\:=\:3$

$\implies\frac{48}{16}\:=\:3$

$\implies\:3\:=\:3$

______________________

Case (2) :- when R1 and R2 in series

Total resistance will be

$\implies\:(R1+R2)\:=\:16$

keep value of R1 and R2 ,

$\implies\:(12+4)\:=\:16$

$\implies\:(16)\:=\:16$

here, both case are satisfies ,

Hence, we can say that value of resistance are correct .

_____________________

Answer 2

(a)......Various branches of science were well developed.

Hope it helps you...

Have a wonderful day ahead